Question

find an equation for a circle with center (-2,6), tangent (touching at one point) to the y - axis. the equation of the circle is \boxed{}. (type an equation. type your answer in standard form. use integers or fractions for any numbers in the equation )

Explanation:

Step1: Recall the standard form of a circle's equation

The standard form of the equation of a circle with center \((h,k)\) and radius \(r\) is \((x - h)^2 + (y - k)^2 = r^2\). Here, the center is \((-2,6)\), so \(h = -2\) and \(k = 6\).

Step2: Determine the radius

The circle is tangent to the \(y\)-axis. The distance from the center \((-2,6)\) to the \(y\)-axis (where \(x = 0\)) is the absolute value of the \(x\)-coordinate of the center. So, the radius \(r = |-2| = 2\).

Step3: Substitute \(h\), \(k\), and \(r\) into the standard form

Substitute \(h = -2\), \(k = 6\), and \(r = 2\) into \((x - h)^2 + (y - k)^2 = r^2\). We get \((x - (-2))^2 + (y - 6)^2 = 2^2\), which simplifies to \((x + 2)^2 + (y - 6)^2 = 4\).

Answer:

\((x + 2)^2 + (y - 6)^2 = 4\)